Problem: $a(n) = -\dfrac{1}{16} \left(2\right)^{n - 1}$ What is the $4^\text{th}$ term in the sequence?
Solution: This is an explicit formula. All we have to do is plug $n=4$ in the formula to find the $4^\text{th}$ term. $\begin{aligned} a({4}) &=-\dfrac{1}{16}(2)^{{4} - 1} \\\\ &= -\dfrac12 \end{aligned}$ The $4^\text{th}$ term is $-\dfrac12$.